1. 1 TOPIC 7: GASES Contents Properties of Gases The Simple Gas Laws The Ideal Gas Equation Gases in Chemical Reactions Mixture of Gases Kinetic-Molecular Theory of Gases Gas Properties Relating to the Kinetic- Molecular Theory Nonideal(Real) Gases

2. 2 PROPERTIES OF GASES Pressure : A force per unit area Force(N) Area(m 2 ) P(Pa) = Liquid Pressure P= g x h x d g: acceleration of gravity h: height of the liquid column d: density

3. 3 BAROMETRIC PRESSURE Standard Atmospheric (Barometric) Pressure 1,00 atm= 760 mmHg, 760 torr 101,325 kPa 1,01325 bar 1013,25 mbar Atmospheric (Barometric) pressure) d Hg = 13,5951 g/cm 3 (0°C) g = 9,80665 m/s 2 Evangelista Torricelli, 1643

4. 4 Manometers Gas pressure equal to barometric pressure Gas pressure greater than barometric pressure Gas pressure less than barometric pressure Measurement of Gas Pressure using an open sided manometer

5. 5 The Simple Gas Laws Boyle’s Law P α 1 V PV = Constant For a fixed amount of gas at a constant temperature, gas volume is inversely proportional to a gas pressure. P 1 V 1 = P 2 V 2

6. 6 Charles’s Law Charles 1787 Gay-Lussac 1802 Volume (mL) V α T V = b T Temperature ( o C) Volume (mL) Temperature (K) The volume of a fixed amount of gas at a constant pressure is directly proportional to the Kelvin(absolute) temperature. Absolute temperature scale: 273,15 o C or 0 K, T(K)= t( o C)+ 273,15 V 1 V 2 T 1 T 2 =

7. 7 Avogadro’s Law Gay-Lussac 1808 Gases react by volumes in the ratio of small whole numbers. Avogadro 1811 Equal volumes of different gases compared at the same temperature and pressure contain equal number of molecules. Equal numbers of molecules of different gases compared at the same temperature and the same pressure occupy equal volumes. V α n or V = c n Standard Conditions (0  C= 273,15 K and 1atm= 760 mm Hg) 1 mol gas = 22,414 L at STP At a constant pressure and temperature :

8. 8 Combination of all Gas Laws: The ideal Gas Equation Boyle’s Law V α 1/P Charles’s Law V α T Avogadro’s LawV α n PV = nRT V α nT P

9. 9 Gas Constant R =R = PV nT = 0,082057 L atm mol -1 K -1 = 8.3145 m 3 Pa mol -1 K -1 PV = nRT = 8,3145 J mol -1 K -1 = 8,3145 m 3 Pa mol -1 K -1

10. 10 Example: What is the volume occupied by 13,7 g Cl 2 (g) sample at 45  C and 745 mm Hg? 1 atm = 760 mmHg; R = 0,08206 L atm /(mol K); Cl:35,5 Solution Practice: What is the pressure exerted by 1,00 x 10 20 molecules of N 2 in a 350 ml volume of container at 175  C ?

11. 11 General Gas Equation R =R = = P2V2P2V2 n2T2n2T2 P1V1P1V1 n1T1n1T1 = PsVsPsVs nsTsnsTs PiViPiVi niTiniTi We often apply it in cases in which one or two of the gas properties are held constant and we can simplify the equation by eliminating these constants In the cases of the comparison of two gases, General Gas Equation must be used. In other cases the ideal gas equation is rather relevant.

12. 12 Applications of the Ideal Gas Equations Is the amount of gas given or asked? Use the General Gas Equation by comparing the initial and final conditions PiVi = PsVs Ti Ts Vi=Vs Pi = Ps Ti Ts No Yes If the mass of gas is constant use the Ideal Gas Equation PV=nRT If the mass of gas is variable use the General Gas Equation. PiVi = PsVs niTi nsTs Boyle’s Law PiVi = PsVs Ti=Ts Vi = Vs Ti Ts Pi = Ps

13. 13 Molar Mass Determination Propylene is an important commercial chemical. It is used in the synthesis of other organic chemicals and in plastic production. A glass vessel weighs 40,1305 g, when clean,dry and evacuated ; 138,2410 g when filled with water at 25°C (density of water δ= 0,9970 g/cm 3 ) and 40.2959 g when filled with propylene gas at 740,3 mm Hg and 24,0°C. What is the molar mass of propylene? Strategy: Find out V ves, m gas ; Use the Ideal Gas Equation V ves = m H 2 O / d H 2 O = (138,2410 g – 40,1305 g) / (0,9970 g cm -3 ) m gas : = 0,1654 g m gas = m - m empty = (40,2959 g – 40,1305 g) = 98,41 cm 3 = 0,09841 L

14. 14 Ideal Gas Equation: PV = nRT PV = m M RT M = m PV RT M = (0,9741 atm)(0,09841 L) (0,6145 g)(0,08206 L atm mol -1 K -1 )(297,2 K) M = 42,08 g/mol

15. 15 Gas Densities PV =PV = m M RT MP RT V m = d = PV = nRT ve d = m V, n = m M Gas densities differ from solid and liquid densities in two important ways: 1- Gas densities depend strongly on temperature and pressure; increasing as the gas pressure increases, decreasing as the temperature increases. Densities of liquids and solids also depend somewhat on temperature, but they depend far less on pressure 2- The density of a gas is directly proportional to its molar mass. No simple realtionship exists between density and molar mass for liquids and solids.

16. 16 Gases in Chemical Reactions Use the stoichiometric factors to relate the amount of a gas to amounts of other reactants or products. Use the ideal gas equation to relate the amount of gas to volume,temperature and pressure. Law of combining volumes can be modified with the other laws

17. 17 Law of Combining Volumes At times, if the reactants and/or products involved in a stoichiometric calculation are gases, we can use a particularly simple approach: 2NO(g) + O 2 (g) 2NO 2 (g) 2 mol NO + 1 mol O 2 (g) 2 mol NO 2 (g) Suppose the gases are compared at the same T and P, in this case one mol of gas occupies a particular volume 1V, 2 mol of gas 2V and 3 mol of gas 3V of liters. 2NO(g) + O 2 (g) 2NO 2 (g) 2 L NO(g) + 1 L O 2 (g) 2 L NO 2 (g)

18. 18 Example: The decomposition of sodium azide, NaN 3, produces N 2 (g). Together with the necessary devices to initiate the reaction and trap the sodium metal formed, this reaction is used in air bag safety systems. What volume of N 2 (g) measured at 735 mm Hg and 26°C is produced when 70,0 g NaN 3 is decomposed? 2 NaN 3 (s) → 2 Na(l) + 3 N 2 (g) Calculate the mole of N 2 : Calculate the volume of N 2 n N 2 = 70 g N 3 x 1 mol NaN 3 65,01 g N 3 /mol N 3 x 3 mol N 2 2 mol NaN 3 = 1,62 mol N 2 = 41,1 L P nRT V = = (735 mm Hg) (1,62 mol)(0,08206 L atm mol -1 K -1 )(299 K) 760 mm Hg 1.00 atm

19. 19 Mixtures of Gases Partial pressure –Dalton’s law of partial pressures: The total pressure of a mixture of gases is the sum of partial pressures of the components of the mixture. P top = P a + P b + P c + … Simple gas laws and ideal gas equation are applicable to a mixture of gases such as air. A simple approach to working with gas mixtures is to use n tot, total amount in moles

20. 20 Dalton’s Law of Partial Pressures P tot = P a + P b +… V a = n a RT/P tot ve V tot = V a + V b +… VaVa V tot n a RT/P tot n top RT/P tot = = nana n tot PaPa P tot n a RT/V tot n tot RT/V tot = = nana n tot nana =  a ( Mole Fraction ) Remember

21. 21 Kinetic-Molecular Theory of Gases A gas is composed of a very large number of extremely small particles in constant, random, straight line motion. Molecules of gas are seperated by great distances(The molecules are treated as if they have a mass but no volume,so called point masses. Molecules collide with one another and with the walls of their container very rapidly. There are assumed to be no forces between molecules. That is each molecule acts independently of all others.In a collection of molecules at constant temperature the total energy remains constant.

22. 22 Gas Properties relating to the Kinetic Molecular Theory Diffusion - The migration of molecules of different substances as a result of random molecular motion. Effusion The escape of gas molecules from their container through a tiny orifice or pin hole.

23. 23 Graham’s Law Graham’s Law: The rates of effusion of two different gases are inversely proportional to the square roots of their molecular mass. Graham’s Law applies only if certain conditions are met. For effusion the gas pressure must be very low, not as a jet of gas. Rate of effusion, u rms = 3 RT=N A mv 2,Note that the product N A m represents the mass of 1 mol of molecules, The molar mass M,so: E K = 3/2RT

24. 24 Effusion

25. 25 Example

26. 26 Real Gases Compressibility factor: PV/nRT = 1 PV= nRT – Ideal gas behaviour –PV/nRT > 1 – At very high pressures –PV/nRT < 1 – Where intermolecular forces of attraction exist

27. 27 Real Gases